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Flux through a surface F=3xi+3yj+zk

Short answer

Close the paraboloid with the base disc and apply Gauss theorem. div F = 7, volume under z=9−x²−y² is 81π/2, total flux 567π/2. Base flux is zero, so flux through S = 567π/2.

If S is the part of the surface z = 9 − x² − y² with z ≥ 0 and F = 3x i + 3y j + z k, find the flux of F through S.

S is the open paraboloid cap z = 9 − x² − y² (z ≥ 0). Its boundary is the circle x² + y² = 9 in the plane z = 0. Close the surface with the disc S₁: x² + y² ≤ 9, z = 0, and use Gauss' theorem.

F = 3x i + 3y j + z k
div F = 3 + 3 + 1 = 7

Volume under the paraboloid

V = ∬ (9 − r²) r dr dθ = 2π [9r²/2 − r⁴/4]₀³ = 2π (81/2 − 81/4) = 81π/2

Total flux through the closed surface

∬_(S+S₁) F·n dS = 7V = 7 × 81π/2 = 567π/2

Flux through the base

On S₁: z = 0, n = −k, so F·n = −z = 0. Flux through S₁ = 0.

Flux through S = 567π/2 − 0

Answer: Flux through S = 567π/2 ≈ 890.6.

✓ Answer checked by Claude Opus 5.5 (high effort)

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